If the sum of n terms of an A.P is n2 then its common difference is ______

Vijay Mukati
6 years ago
common difference, d = 2(n-a)/(n-1)
deepak
15 Points
4 years ago
Ans is 2 bcz AP is 1,3,5,7,9,11 here u can see that sum of first 2 terms is 4 thatis 2 sq. , sum of first 3 tems is 9 that is 3 sq. And so on
AMRITA GHOSH
11 Points
4 years ago
n/2 [ 2a +(n-1)d] = nOn solving ,2n - 2a = (n-1)d=> d = 2(n-a)/(n-1)Now, when n=1,d = 2(2-a)/(2-1) = 4 - 2aWhen n=2,d= 2(3-a)/(3-1) = 3-aWe know, that d is constant in an AP series. Hence, d = 4-2a = 3-a=> 2a- a= 4 - 3=> a = 1=> d = 3 - a = 3 - 1 = 2
Samyak Jain
333 Points
4 years ago
We know that sum of n terms of an AP is
(n/2)[2a + (n – 1)d] , where a is the first term & d is the common difference of the AP.
It is given that sum of an AP is n.
So n2  = (n/2)[2a + (n – 1)d]
i.e. n = a + (n – 1)d / 2
Put n = 1  ,  1 = a
Put n = 2  ,  2 = a + d / 2   i.e.  2 = 1 + d/2
$\dpi{80} \therefore$ d = 2
The common difference of the given AP is 2.